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Chapter 16: Problem 94

Calculate the \(\mathrm{pH}\) of a solution obtained by mixing \(35.0 \mathrm{~mL}\) of \(0.15 M\) acetic acid with \(25.0 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) sodium acetate.

Short Answer

Expert verified
The pH of the solution is approximately 4.44.

Step by step solution

01

Identify Key Components

First, identify the components involved in this problem: we have a weak acid, acetic acid, and its conjugate base, sodium acetate. This combination forms a buffer solution.
02

Understand the Buffer Solution

A buffer solution resists changes in pH when small amounts of acid or base are added. It consists of a weak acid and its conjugate base. The Henderson-Hasselbalch equation will be used: \[ \mathrm{pH} = \mathrm{pK_a} + \log\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right) \]where \(\mathrm{pK_a}\) of acetic acid is 4.76.
03

Calculate Initial Moles of Acid and Base

Determine the initial moles of acetic acid and sodium acetate:- Moles of acetic acid = volume \(\times\) concentration = \( 35.0 \times 10^{-3} \times 0.15 = 5.25 \times 10^{-3} \) moles.- Moles of sodium acetate = volume \(\times\) concentration = \( 25.0 \times 10^{-3} \times 0.10 = 2.5 \times 10^{-3} \) moles.
04

Calculate Total Volume of the Solution

Calculate the total volume of the solution after mixing:\( V_{\text{total}} = 35.0 \text{ mL} + 25.0 \text{ mL} = 60.0 \text{ mL} \) or \(60.0 \times 10^{-3} \text{ L} \).
05

Calculate Concentrations of Acid and Base in the Mixture

Concentration of acetic acid in the mixture:\[ [\mathrm{HA}] = \frac{5.25 \times 10^{-3}}{60.0 \times 10^{-3}} = 0.0875 \text{ M} \]Concentration of acetate ion in the mixture:\[ [\mathrm{A^-}] = \frac{2.5 \times 10^{-3}}{60.0 \times 10^{-3}} = 0.0417 \text{ M} \]
06

Use the Henderson-Hasselbalch Equation

Apply the concentrations to the Henderson-Hasselbalch equation:\[ \mathrm{pH} = 4.76 + \log\left(\frac{0.0417}{0.0875}\right) \]
07

Calculate the Logarithm

Compute the logarithmic term:\[ \log\left(\frac{0.0417}{0.0875}\right) = \log(0.476) \approx -0.323 \]
08

Determine the Final pH

Add the logarithmic result to the \(\mathrm{pK_a}\):\[ \mathrm{pH} = 4.76 - 0.323 = 4.437 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acetic Acid
Acetic acid, known chemically as \( \mathrm{CH_3COOH} \), is a weak organic acid with a distinctive sour taste and pungent smell. It is a vital component in everyday substances like vinegar, where it contributes to the tangy flavor. As a weak acid, acetic acid does not completely dissociate in water, which means only a small fraction of its molecules ionize to release hydrogen ions. This partial ionization is key to its behavior in buffer solutions, where it can interact with its conjugate base to resist changes in pH. When acetic acid is combined with soluble salts that contain its conjugate base, such as sodium acetate (\( \mathrm{CH_3COONa} \)), the resulting mixture can maintain a steady pH even when external acids or bases are introduced. This quality is crucial in many biological and chemical systems.
pH Calculation
Calculating \( \mathrm{pH} \) essentially measures the acidity or basicity of a solution. The \( \mathrm{pH} \) scale, ranging from 0 to 14, indicates how acidic or basic a solution is, with lower values signifying higher acidity. The calculation involves determining the concentration of hydrogen ions (\( [\mathrm{H^+}] \)) in the solution. The formula for pH is:
  • \( \mathrm{pH} = -\log([\mathrm{H^+}]) \)
In the case of acetic acid and sodium acetate forming a buffer, we use the Henderson-Hasselbalch equation to find the pH because it relates pH directly to the concentrations of the acid and its conjugate base. The log term in the equation uses the ratio of these concentrations, helping us determine how shifts in these values can affect the overall pH. Understanding pH calculation through this context is important in fields such as chemistry and biology, where maintaining specific pH levels is crucial for experimental accuracy and biological processes.
Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation is a mathematical tool used to estimate the pH of buffer solutions. It provides a straightforward means to relate the pH, the concentration of an acid (\([\mathrm{HA}]\)), its conjugate base (\([\mathrm{A^-}]\)), and the acid dissociation constant \( \mathrm{pK_a} \). The equation is expressed as:
  • \[ \mathrm{pH} = \mathrm{pK_a} + \log\left(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\right) \]
This equation simplifies the process of finding the pH of a buffer solution by using known values. By plugging in the concentrations from our mixture, we can directly calculate the pH. The Henderson-Hasselbalch equation assumes that the concentration of the acid and its conjugate base won't change significantly due to dissociation. It is extremely useful in biochemical applications, where precise pH conditions are necessary, such as in enzyme reactions and drug stability studies. This equation aids in understanding how buffers work and why they are crucial for maintaining equilibrium in chemical systems.

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Most popular questions from this chapter

Hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S}\), is a very weak diprotic acid. In a \(0.10 M\) solution of this acid, which of the following would you expect to find in the highest concentration? a. \(\mathrm{H}_{2} \mathrm{~S}\) b. \(\mathrm{H}_{3} \mathrm{O}^{+}\) c. \(\mathrm{HS}^{-}\) d. \(\mathrm{S}^{2-}\) e. \(\mathrm{OH}^{-}\)

A solution of acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\), on a laboratory shelf was of undetermined concentration. If the pH of the solution was found to be \(2.68\), what was the concentration of the acetic acid?

A 0.288-g sample of an unknown monoprotic organic acid is dissolved in water and titrated with a \(0.115 M\) sodium hydroxide solution. After the addition of \(17.54 \mathrm{~mL}\) of base, a \(\mathrm{pH}\) of \(4.92\) is recorded. The equivalence point is reached when a total of \(33.83 \mathrm{~mL}\) of \(\mathrm{NaOH}\) is added. a. What is the molar mass of the organic acid? b. What is the \(K_{a}\) value for the acid? The \(K_{a}\) value could have been determined very easily if a pH measurement had been made after the addition of \(16.92 \mathrm{~mL}\) of \(\mathrm{NaOH}\). Why?

How can you account for the fact that normal rain is slightly acidic?

a. For each of the following salts, write the reaction that occurs when it dissociates in water: \(\mathrm{NaCl}(s), \mathrm{NaCN}(s)\), \(\mathrm{KClO}_{2}(s), \mathrm{NH}_{4} \mathrm{NO}_{3}(s), \operatorname{KBr}(a q)\), and \(\mathrm{NaF}(s)\) b. Consider each of the reactions that you wrote above, and identify the aqueous ions that could be proton donors (acids) or proton acceptors (bases). Briefly explain how you decided which ions to choose. c. For each of the acids and bases that you identified in part b, write the chemical reaction it can undergo in aqueous solution (its reaction with water). d. Are there any reactions that you have written above that you anticipate will occur to such an extent that the \(\mathrm{pH}\) of the solution will be affected? As part of your answer, be sure to explain how you decided. e. Assume that in each case above, \(0.01 \mathrm{~mol}\) of the salt was dissolved in enough water at \(25^{\circ} \mathrm{C}\) to make \(1.0 \mathrm{~L}\) of solution. In each case, what additional information would you need in order to calculate the \(\mathrm{pH}\) ? If there are cases where no additional information is required, be sure to state that as well. f. Say you take \(0.01\) mol of \(\mathrm{NH}_{4} \mathrm{CN}\) and dissolve it in enough water at \(25^{\circ} \mathrm{C}\) to make \(1.0 \mathrm{~L}\) of solution. Using chemical reactions and words, explain how you would go about determining what effect this salt will have on the \(\mathrm{pH}\) of the solution. Be sure to list any additional information you would need to arrive at an answer.
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